5a) Find the values of p for which the equation x^2 + px + 8 = p has one positive and one negative root [p > 8]

I have found the determinant to be p^2 + 4p - 32

b) Find the positive value of k for which the equation 6x^2 - 2kx + k = 0 has 2 real roots, one being thrice the other

I have found the determinant to be k^2 - 6k

c) Given that A and B are the roots of the euqation x^2 = 3x + 5, find the value of 1/A^2 + 1/B^2 and show that A^4 = 57A + 70

I have found the answer to be 19/25, but I've no idea how to show the equation

Thanks a lot

2. For parts (a) and (b), the determinants must be greater than 0 for the quadratic equations in x to have two real solutions. Therefore, it's a matter of solving the inequalities $\displaystyle p^2 + 4p - 32 > 0$ and $\displaystyle k^2 - 6k > 0$.

3. Originally Posted by Drdj
5a) Find the values of p for which the equation x^2 + px + 8 = p has one positive and one negative root [p > 8]

I have found the determinant to be p^2 + 4p - 32 <--- this term is called discriminant

...
1. To solve the equation for x use the quadratic formula:

$\displaystyle x^2+px+8-p=0$

$\displaystyle x = \dfrac{- p + \sqrt{p^2 +4p-32}}2~\vee~ x=\dfrac{- p - \sqrt{p^2 + 4p-32}}2$

2. Solve the inequalities for p:

$\displaystyle \dfrac{- p + \sqrt{p^2 +4p-32}}2 > 0 ~\wedge~ \dfrac{- p - \sqrt{p^2 +4p-32}}2 <0$

$\displaystyle \sqrt{p^2 +4p-32} > p ~\wedge~ - \sqrt{p^2 +4p-32} < p$

$\displaystyle p\le-8\vee p >8~~~~~~\wedge~~~~~~p\ge 4~\implies~p > 8$

3. Now solve the inequalities for p:

$\displaystyle \dfrac{- p + \sqrt{p^2 +4p-32}}2 < 0 ~\wedge~ \dfrac{- p - \sqrt{p^2 +4p-32}}2 > 0$

You should come out with $\displaystyle p \le -8$

4. Originally Posted by Drdj
c) Given that A and B are the roots of the euqation x^2 = 3x + 5, find the value of 1/A^2 + 1/B^2 and show that A^4 = 57A + 70
$\displaystyle x^2 = 3x+5 \iff x^2-3x-5 = 0$.
By Vieta's, $\displaystyle a+b = 3$ and $\displaystyle ab = -5$, and

$\displaystyle \displaystyle \frac{1}{a^2}+\frac{1}{b^2} = \frac{b^2+a^2}{a^2b^2} = \frac{(a+b)^2-2ab}{(ab)^2}.$

5. Thanks 'TheCoffeeMachine'
Can you help me with the 2nd part - show that A^4 = 57A + 70 ?
Thanks

6. Thanks for your help, earboth, though I don't quite understand your method
Is there an alternative way, or could you elaborate?
Apologies

I'm sorry, I'm not sure how the second part follows. Perhaps they want you to find $\displaystyle a$ and $\displaystyle b$ explicitly first?